PS: before today I didn't know that this "thing" was called "Maximum power =
point tracking",
so I took a look at the Wikipedia page for it:

http://en.wikipedia.org/wiki/Maximum_power_point_tracking

And I realized that the technique/algorithm I described is not optimal for =
solar panels, but
for purely resistive sources (like the one I was using when I played/experi=
mented with it).

For the OP: take a look at the above Wikipedia page.. it's very interesting=
, and talks
specifically about solar panels, which have a very non-linear current/volta=
ge relationship.

With kind regards,
Mario



>At 11.50 2011.09.17, you wrote:
>>Generaly...
>>
>>If you hava a battery it has an internal resistance. I think you want
>>to pull out the maximum power from this power source. In this case
>>your external resistor sould be equal with the internal resistance. If
>>you increase the value of the external resistance, the current will
>>drop on it ( altough the voltage will be higher ). If you decrease the
>>value of the external resistance, the voltage will drop on it (
>>altough the current will be higher ). So if you multiply the voltage
>>with current the result will be the actual power on the external
>>component.
>>
>>For example:
>>U battery =3D 600V
>>R internal =3D 50 Ohm
>>R external =3D 50 Ohm
>>
>>I =3D U / ( Ri + Re ) =3D 6 Amper
>>U ext =3D 300V
>>P ext =3D 300V * 6 Amp =3D 1800 Watt
>>
>>------
>>
>>U battery =3D 600V
>>R internal =3D 50 Ohm
>>R external =3D 30 Ohm
>>
>>I =3D U / ( Ri + Re ) =3D 7.5 Amper
>>U ext =3D 600V * 30 / ( 50 + 30 ) =3D 225V
>>P ext =3D 225V * 7.5 Amp =3D 1687.5 Watt
>>
>>------
>>
>>U battery =3D 600V
>>R internal =3D 50 Ohm
>>R external =3D 100 Ohm
>>
>>I =3D U / ( Ri + Re ) =3D 4 Amper
>>U ext =3D 600V * 100 / ( 50 + 100 ) =3D 400V
>>P ext =3D 400V * 4 Amp =3D 1600 Watt
>>
>>If you have a solar panel, you can't know the exact value of the
>>internal resistance, because sunlight doesn't come countinuesly and
>>the clouds disturb the sunlight also. So you have to found the maximum
>>point where you can pull the maximum energy from the solar cell or
>>other power source because it depends on the illumination.
>
>And how do you do it? So far that's the best approach I thought, but if
>someone has a better idea, I'd like to hear it!
>
>In my PIC based SMPS boost circuit, I unload the source (panel in this
>case) until the voltage stabilizes, and then save the value into a registe=
r.
>Then I turn on the switch and the inductor starts charging, thus current
>increases. When the source voltage becomes half the open voltage I have
>previously recorded, that's the optimum current I have to sink from the
>source, thus I let e.g. 50mA more for it, then turn off the transistor
>switch, until the current decreses to e.g. 50mA less than that optimum
>value, then switch the transistor ON, and continue the loop, until the
>output capacitors are fully charged, then I stop it. About every second,
>I force a return to the outer loop even if the capacitor is not fully
>charged, i.e. I measure input voltage at open (not loaded) conditions.
>I love to design SMPS using MPU's, never ever used a switching IC in my
>life, although of course there are situations (mostly battery-powered
>small devices) where it would make perfectly sense to use one.. but for
>high power stuff, I prefer to make my own algoritms and use PIC's.
>
>Is there a better way to implement maximum power point tracking?
>
>With kind regards,
>Mario
>
>
>
>>
>>--=20
>>~~~~~~~~~~~~~~~~
>>http://galzsolt.zzl.org

--=20
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